The same can be written as
h ¼
2H
3
þ
K
3
þ
L
3
k ¼ À
H
3
þ
K
3
þ
L
3
l ¼ À
H
3
À
2K
3
þ
L
3
ð5:15Þ
Example 3 Find the equivalent HCP planes for the following rhombohedral
planes:ð10 1Þ) and (200).
Solution: Given: Rhombohedral planes: ð10 1Þ) and (200).
Let us take them one by one.
Case I: Rhombohedral plane: (hkl) ð10 1Þ).
Substituting the values of h, k and l in Eq. 5.11, we obtain (HKIL) ð11 20Þ for
HCP. Again, substituting the values of H, K and L in Eq. 5.15, we obtain (hkl)
ð10 1Þ for rhombohedron, this is the same indices with which we started. This
confirms the validity of Eqs. 5.11 and 5.15.
Case II: Rhombohedral plane: (hkl) (200).
A similar operation with the given h, k and l values will provide us (HKIL)
ð20 22Þ for HCP. Again, substituting the values of h, k and l in Eq. 5.15, we obtain
(hkl) (200) for rhombohedron. This confirms the validity of Eqs. 5.11 and 5.15.
Using these equations, a one to one correspondence of other Miller indices can be
obtained.
4. Trigonal and simple Hexagonal
The axial relationships between the translation vectors of trigonal and primitive
(simple) hexagonal unit cells are shown in Fig. 5.4. Let us consider a crystal plane
which is referred to as (HKL) in trigonal system of axes A 1 , A 2 , A 3 and (hkil) in
hexagonal system of axes a 1 , a 2 , c.
From Fig. 5.4, we can write the following axial relationships:
A 1 ¼ a 1 þ c /3, A 2 ¼ a 2 + c/3, A 3 ¼Àa 1 À a 2 + c/3
Based on above relationships, we can express trigonal indices in terms of the
primitive hexagonal indices through the following transformation equations
H ¼ 1h þ 0k þ 1l/3
K ¼ 0h þ 1k þ 1l/3
L ¼ À 1h À 1k þ 1l/3
ð5:16Þ
The matrix form of these equations is
186
5 Unit Cell Transformations